chapter5 notes me311 f14 pjf
TRANSCRIPT
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MASS AND ENERGY ANALYSIS OF
CONTROL VOLUMES
Florio
Cengel
10-2014
2
Objectives• Develop the conservation of mass principle.
• Apply the conservation of mass principle to various control volumes
systems.
• Apply the first law of thermodynamics as the statement of the
conservation of energy principle to control volumes.
• Identify the energy carried by a fluid stream crossing a control surface as
the sum of internal energy, flow work Pv, kinetic energy, and potential
energy of the fluid and to relate the combination of the internal energy and
the flow work to the property enthalpy.
• Solve energy balance problems for common steady-flow devices such as
nozzles, compressors, turbines, throttling valves, mixers, heaters, and
heat exchangers.
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CONSERVATION OF MASS
3
Mass is conserved even during chemical reactions , in addition there is
conservation of the atomic species..
Conservation of mass: Mass, like energy, is a conserved property, and it cannot becreated or destroyed during a process only changed in form and/or location.
Closed m ass systems : The mass of the system remain constant during a process.
Control volumes : Mass can cross the boundaries, and thus the identity of the matterwithin the region changes; so we must keep track of the amount of mass entering andleaving the control volume.
Mass m and energy E can be converted to each other according to
where c is the speed of light in a vacuum, which is c = 2.9979 108 m/s.
For most of our problems, the mass change due to energy change is negligible.
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Mass and Volume Flow Rates
4
The volume flow rate is the volume
of fluid flowing through a cross
section per unit time.
Definition of
average velocityMass flow
rate
Volume flow rate-define average velocity
dA
avg
avg avg
m Vol
V A
Ac
Vavg is the uniform velocity necessary to
produce the same volume flow rate.
Vavg =VOLdot/Ac
PdA
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Conservation of Total Mass Principle
5
The conservation of mass principle for a control volume : The net mass transfer to or
from a control volume during a time interval t is equal to the net change (increase or
decrease) in the total mass within the control volume during t . M is the amount of mass
General conservation of mass 0 place holder for DM/Dtfmsof the fms momentarily occuping the cv at time t-
Reynolds- Transport eq.
General-exact-conservation of mass in rate form
or
Atomic species or molecular
species form is:Volume flow conservation for
steady fixed cv with
incompressible flow.
( )V N dA
∆
∆
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Mass Balance for Steady-Flow Processes
6
Conservation of mass principle for a two-
inlet –one-outlet steady-flow system.
During a steady-flow (rate of exchanges), steady state within -fixed CV process,
the total amount of mass contained within a control volume does not change with time
(mCV = constant).
Then the conservation of mass principle requires that the total rate of mass entering
a control volume must equal the total rate of mass leaving it.For steady-flow processes, we are interested in
the amount of mass flowing per unit time, that is,
the mass flow rate.
Multiple inlets
and exits
Single
streamMany engineering devices such as nozzles,
diffusers, turbines, compressors, and pumps
involve a single stream (only one inlet and one
outlet).
21
3
For a single stream, 1 in, 2 out, and using bulk properties
Gt 37
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Images
• video 1
• video 2
• Video 3
• Video 4
• Video 5
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8
/kgm001017.0 C65
MPa7 31
1
1
v
T
P 6 MPa, 450C
80 m/s
Steam7 MPa65C
5-8 Water flows through the tubes of a boiler. The velocity and volume flow rate
of the water at the inlet are to be determined. The internal diameter is 0.130 m and the water enters at 7MPa,65 C and exits at 6MPa 450 C, 80 m/s.
Assumpt ions Flow through the boiler is steady., and cv is fixed in size and shape
Propert ies The specific volumes of water at the inlet and exit are (Tables A-6 and A-7)
/kgm05217.0 C450
MPa6 32
2
2
v
T
P
Analysis The cross-sectional area of the tube is
222
m01327.04
m)13.0(
4
D Ac
The mass flow rate through the tube is same at the inlet and exit. It may be determined from exit data to be
kg/s20.35/kgm0.05217
m/s)80)(m01327.0(3
2
2
2 v
V Am c
The average water velocity at the inlet is then3
11 2
(20.35 kg/s)(0.001017 m /kg)/
0.01327 mc
mV Vol A
A
v 1.560 m/ s
The volumetric flow rate at the inlet is
2
1 1(0.01327 m )(1.560 m/s)
c A V 30.0207 m / sVol
Rtn 6
Qdot
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http://localhost/var/www/apps/conversion/tmp/scratch_4/FIG04_03.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_1.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_15.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_17.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_19.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_19.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_19.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_19.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_17.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_15.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_1.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_1.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_1.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG04_03.MOV
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9
2 3
room
3 3
room
define air vol change rate = [(2.7 m)(200 m ) 540 m /room
(540 m )(0.35/h) 189 m / h 189,0
/
00 L/
]
h
room time
3150 L / min
Vol
Vol Vol
House200 m2
0.35 ACH
Ex3- The minimum fresh air requirements of a residential building is specified
to be 0.35 air volume changes per hour (volume changes), or 35% of the air in the room is to be replaced every
hour. The room is assumed to be 2.7m high and 200 m2 floor area.. The size of the fan capacity in L/min that
needs to be installed and the diameter of the duct are to be determined if the average air speed is 6 m/s.
Analysis Assume air pressure change is small so air density constant ∆P IS SMALL relative to P Steady flow
volume flow in is equal to out. The volume of the building and the required minimum volume flow rate of fresh air
are
The volume flow rate of fresh air can be expressed as
2 2( / 4) ; VD = const so Vmax, Dmin and vise versaVA V D Vol
Solving for the diameter D and substituting,
34 4(189 / 3600 m /s)
(6 m/s) D
V 0.106 m
Vol
Therefore, the diameter of the fresh air duct should be at least 10.6 cm if the velocity of air
is not to exceed 6 m/s .
Return to
7
outflow
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Control volume-Control Mass-simple- in time dt a small element of mass enters the
resulting balance equations
• Fixed mass form to the control volume form
• dM)cv = dm)in The total workby is dWother,SHAFT,SHEAR,ELECT - P in [v//v]*dvol in +P out [v/v]*dvol out ; Tointroduce the mass into the CV an effort must be expended
• [δQ- δ Wother+ (P*dvolin-Pvdmout) + e*dmin - e*dmiout)=dE)cv] - in time dt• Where Pdvol is work done by normal forces at flow port in
• dmin = dVol/v)in = [A*Vrel,n, /v]*dt
10
M,E
e*dmin through
an imaginary
mesh
Wflowon =P*d(volin)
dQdot through the
entire surface,
including flow portsurface(s)
dWdot
through the
entire surface
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FLOW WORK for A CONTINUOUSLY FLOWING FLUID
11Schematic for flow work.
Flow work-- P* voldot, : The work (or energy transfer)
required to push the mass into (on) or out (by) of the
control volume or the work done by normal stresses atthe flow ports. This work is necessary for maintaining a
continuous flow through a control volume.
In the absence of acceleration, the force
applied on a fluid by a piston is equal to the
force applied on the piston by the fluid.10/14/2014 Dr. Florio
Total Energy of a Flowing Fluid= stored +flow work
12
The total energy consists of three parts for a non-flowing fluid and for a flowing fluid energy transfer
associated with the mass consists four parts flow work and the 3 of e.
h = u + Pv
The transfer of energyat a flow port-flow work-is automatically taken
care of by enthalpy. The
product of properties
Pv represents the flow
work per unit of mass
only at the flow port
otherwise it is a
product of properties.
dPv pdV vdP
The quantity is called the methalpy , the total enthalpy
or transfer energy per unit of mass
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Energy Transport by Mass-bulk motion
iim
13
At any instant the product
is the energy transported into control
volume by mass per unit time.
When the kinetic and potential energies of a
fluid stream are negligible the energy carried
in
Exact form when the properties of the
mass at each inlet or exit change with
time as well as over the cross section
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Steady State
• 1. If the states within do not change with time, the system is said tobe in a steady state.-[Black Box]
• 2. Steady flow implies all the rate exchange quantities, mdot ,Wdot,Q dot do not vary with time
• 3. For a control volume whose boundary does not change and is ina steady state, is also acting under steady flow conditions. Thus theaccumulation of all extensive properties are zero.
• 4. It is sometimes convenient to define specific work and heatquantities using the reference mass flow rate
• 5.
14
/ and w=W/mref ref
q Q m
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ENERGY ANALYSIS OF STEADY-FLOW-FIXED- SYSTEMS- any
extensive within cv is constant
15
Under steady-flow, fixed conditions, the mass
and energy contents of a control volume remain
constant.
Under steady-flow, fixed,
conditions, the fluid properties at
an inlet or exit remain constant
(do not change with time).
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Mass and Energy balances for a steady-
flow process
16
A water heater
in steady
operation.
Mass
balance
Energy
balance
opensystem
OR
,fixed)
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17
Under steady operation,shaft work, shear and
electrical work are the only
forms of work a simple
compressible system may
involve.
Energy balance relations with sign conventions (i.e.,
heat input and work output are positive)
when kinetic and potential energy
changes are negligible
Caution on energy units
Eq below, path must be known
out
rev shaft
in
w vdP ke pe for a fms
rev rev
rev rev
q h vdP
while
q u Pdv
Most handy form
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Relative importance- Steady Flow-SF
• The relative importance of various terms differ indifferent energy conversion applications.
• In some applications some terms are usually smallor negligible (but not all the times). Level of ke
• Energy units must be consistent
• Best in applications to first write the full steadyflow equations and then eliminate the negligibleterms.
• Customize in light of the circumstances andobjectives of the particular application.
18
2 2 / 2 [45 / 2]/1000 =1 k / ] If V J kg
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Series of -Analysis –applications
Not internal design use to Synthesized
• The following is a series of applications involving commoncomponents that will be packaged to form components of a system.
• Energy conversion devices, M=mech E type;i.e. work to pressure,ke,or pe
• M-M, M+T-M, M+T- [M+T]
• Rev compressible, Mechanical coupled to thermal
• Irreversible for M converted to thermal
• Synthesized
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Nozzles and Diffusers
Shapes for subsonic flows-
converging-Nozzles and diverging-
diffusers are shaped(5-7 degrees)
so that they cause large changes in
fluid velocities and thus kinetic
energies.
Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft, and even
garden hoses P/ρ to V2 /2 or vise-versa.
Devices to alter kinetic energies by means of
geometry changes recall ke can be
completely converted to mech work
A nozzle is a device that increases the kinetic
energy of a fluid at by the transformation of
the pressure E i.e. pressure energy. For a
compressible substance, there is a coupling
between T and P.
The cross-sectional area of a nozzle
decreases in the flow direction for subsonic
flows and increases for supersonic flows. The
reverse is true for diffusers.
A diffuser is a device that increases the
pressure of a fluid by the transformation of the
ke.Energy balance for an
approximate adiabatic
nozzle or diffuser :
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Alternate formulation- Nozzle or diffuser• Convert mechanical and thermal to kinetic, or
vise-versa in the acceleration or decelerationof the flow with no shaft or boundary work:
• For nozzles or diffusers with bulk state1 in
and bulk state2 out is
• :Using Conservation of mass and EE,assuming steady and flows
• q+ h1+V12 / 2J=h2+V2
2 /2J E
• Using the usual assumptions o f steady flow,small potential energy change and adiabatic.J is an energy conversion factor, mech E tothermal measures
• For a general reversible process-1D-steady,
• h2-h1= ∫v dP ****+ qrev
• For incompressible, rev, this is ∆h = v ∆ P+qrev
• Bernouli’s- internally reversible
• ∫v dP +V22 /2 -V1
2 /2=0
• Problem
21
Eq 1
Eq 2
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22
kJ/kg3196.7
/kgm0.057838
C004
MPa5
1
31
1
1
hT
P v
Steam1 2
120 kJ/sHeat loss
Ex 3 Steam at 5 Mpa, 400 C is accelerated in a nozzle from a velocity of 80 m/s
and leaves at 2 Mpa, 300 C. The nozzle inlet area is 50 cm2 and heat is lost at
a rate of 120 kW The mass flow rate, the exit velocity, and the exit area of
the nozzle are to be determined.
Assumpt ions 1 This is a steady-flow process since there is no change with time.
2 Potential energy changes are negligible. 3 There are no work interactions.
Propert ies From the steam tables (Table A-6) speed of soumd –steam=SQRT(kRT)= =24.7*SQRT(T) m/s
and
kJ/kg3024.2
/kgm0.12551
C003
MPa2
2
32
2
2
hT
P v
Analysis (a) There is only one inlet and one exit, and thus
m m m1 2
. The mass flow rate of steam is
kg/s6.92 )m1050)(m/s80(/kgm0.057838
11 24311
1
AV mv
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(b ) Recall m2 /s2 is a J/kg We take nozzle as the system, which is a control volume sincemass crosses the boundary.
The energy balance for this steady-flow system can be expressed in the rate form as
0 (steady+fixed)
in out system
Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies
in out
/ 0 E E dE dt
E E
2 2
2 12 1
(since W pe 0)
2in
V V Q m h h
J
Substituting, the exit velocity of the steam is determined to be
22
222
/sm1000
kJ/kg1
2
m/s)(803196.73024.2kg/s6.916kJ/s120
V
It yields V 2 = 562.7 m/s; speed of sound =591 m/s
(c ) The exit area of the nozzle is determined from
32 2
2 2 2
2 2
6.916 kg/s 0.12551 m /kg1
562.7 m/s
mm V A A V
4 2
15.42 10 mv
v
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• Polytropic nozzle
24
(1 )/
1
Assume we have a reversible polytropic nozzle with
Pv constant and / ( 1)
If ideal gas then P const as well asTv
n
n n
n
vdP n pv n
T const
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EX1-Nozzle problem-
• Using the preceding NOZZLE cv. Fluid is water
• Given: Mdot = 0.1 kg/s, state 1; P=1000 kPa, T= 400 C,• state 2, P=500 kPa, T= 350 C• Find Vexit, Exit Area.
• Assuming SS and fixed, adiabatic and change in pe negligiblethen
• Inlet state 1 V1 =small ;0.0, SH Tables h1= 3263.88 kJ/kg• Exit state 2; h2= 3167.65 kJ/kg, v2 =0.57012 m
3 /kg• EE eq : V2
2 /(2J) = h1-h2 + V12/2J , J=1000J/kJ
• V2= sqrt(2*J*(h1-h2)) =sqrt( 2*1000*(3263.88-3167.65))
• V2= 438.7 m/s (property relation dh= v dP0 )• A=area = mdot*v/V = 0.1*0.57012/438.7=1.3E-04=1.3cm
2
25
2 1m m
2 1m m
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Work in devices• The cv is a fluid cv.
• Assuming steady state, fixedcontrol surface. If heatinformation is unknown,safe to assume small.Applying the 1st law, theequation for the adiabaticwork done by the impelleron the fluid is given by
• Wdotin= =mdot{(h2+ke2+pe2)-(h1+ke1+pe1)}
• Usually flow passage isdesigned so as densityincreases, the flow area
decreases, thus shape andstatement
• Usually ke and pe changesby design are small- but maynot be true for hydraulicsystems
• problem
26
compressor
Pumps-liquids- low dP high flow
Force exerted by impeller, rotor, on the fluid
and moves in direction of the force
2
1
Fby fluid
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Aside: cv is the fluid*****
• Defining specific exchange quantities : q to be Qdot/Mdot, and w tobe Wdot/mdot: where Mdot is usually the net inflow rate. For work
devices and nozzles Qdot is small and Mdot large so q small relative tow .
• For the conditions of ss ,fixed cv , one dimensional and reversibleflow; Using the 2nd Law, h2-h1= 1 2 v*dP ****+ qrev
• the inclusion into the energy equation reduces the energy to
• a. –w)byrev = (ke/J +pe/J+ 1 2 v*dP)• Equation a.(Is 1st +2nd) and equation b.1st
• b. qrev – wbyrev = h + ke/J + pe/J for the steady , fixedconditions.
• These two equations , a and b , are independent.
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EX4 Reversible partial compressor problem cv fluid- Given a non- adiabatic compressor
• NH3 is the flowing fluid. Mdot is 4 kg/s. Inlet state 1 : T= - 20 C, P= 150kPa: outlet 2: T2=80 C, P2=400 kPa.
• Assume the process is a reversible polytropic process.
• Using Pvn =const and assuming ke, pe are small• Wdot rev by = - [ 1 2 v dP] mdot = - ∆h integrating by parts yields• wby)rev = - n(P2*v2-P1*v1) /(n-1)********
• State 1 is defined going to NH3 tables you would find v1=0.79774 m3/kg;
h1=1422.9 kJ/kg, state 2 is defined : v2=0.4216, h2=1636.7 kJ/kg.
• n is required : Using the path eq : Pvn = c
• or n= ln(P2/P1)/ln (v1/v2):
• Natural log yields:n=0.5189, w=-53.1 kJ/kg, ∆h =213.8; qrev=160.7 kJ/kg
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Turbines and
Compressors Turbine drives the electric generator In steam,gas, or hydroelectric power plants.
As the fluid passes through the turbine, work is
done on the blades, which are attached to theshaft. As a result, the shaft rotates, and the
turbine produces work.
Compressors, as well as pumps and fans,
are devices used to increase the pressure of a
fluid. Work is supplied to these devices from an
external source through a rotating shaft.
A fan increases the pressure of a gas slightly
and is mainly used to move a gas Pout./Pin
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SOME STEADY-FLOW ENGINEERING DEVICES
Many engineering devices operate essentially under the same conditions
for long periods of time. The components of a steam power plant (turbines,
compressors, heat exchangers, and pumps), for example, operate nonstop formonths before the system is shut down for maintenance. Therefore, these devices can
be conveniently analyzed as steady-flow devices.
At very high velocities,
even small changes invelocities can cause
significant changes in the
kinetic energy of the fluid.
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32
31 1
1 1
10 MPa 0.029782 m /kg
450 C 3242.4 kJ/kg
P SH
T h
v
STEAMm = 12
kg/s
·
P 1 = 10 MPa
T 1 = 450C
V 1 = 80 m/s
P 2 = 10 kPa
x 2 = 0.92
V 2 = 50 m/s
W ·
Ex-5 Steam expands in a turbine. IF THE MASS FLOW RATE INTO THE TURBINE IS 12 KG/S, The
change in kinetic energy, the power output, and the turbine inlet area are to be determined. See Fig.Assumpt ions 1 This is a steady-flow process since there is no change with time.
2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Propert ies From the steam tables (Tables A-4 through 6)
and
kJ/kg2392.52392.10.92191.8192.0
kPa1022
2
2
fg f h xhh
x
P
Analysis (a) The change in kinetic energy is determined from
2 22 2
2 1
2 2
50 m/s (80 m/s) 1 kJ/kg
2 2 1000 m /s
V V ke
J
1.95 kJ/ kg
m m m1 2
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